M3: newtons law of gravitation

how would I do a?

I see: $F = K/x^2$

but I dont see: $F = - K/x^2$

how would I do a?

how would I do a?

I see: $F = K/x^2$

but I dont see: $F = - K/x^2$

Oh that's wierd, your post changed while I was replying!

Is your only issue the negative sign (or lack thereof)? If so, that's not really important here - it only helps to clarify the direction of the force, its magnitude will be the same, and that's all that they are asking about.

why is it negative anyway?

Really, don't worry about this part. But the idea is that if you take x as a vector pointint out from the centre of the Earth towards the object, then the force has to act the other way, thus the negative sign.

whats going on here? F = -K/x^2

but what is mg = -k/R^2?

They are just saying that if the negative sign has been included in the working, then the method marks can still be scored. Pieces in brackets may or may not be seen in the working, and it doesn't matter if they are there or not for the marks to be awarded.

ok, but whats: mg = -k/R^2?

It's what you get from the previous line when you use F = mg and x = R (that is, the weight of the object at the surface of the Earth).

why do we want the weight of the object at the surface of the earth?

The "F" in the previous line is the gravitational force on the object, that is, it's weight. So when x = R (we are at the surface of the Earth), F = mg (what you would normally use for weight at the surface of the Earth).

The "F" in the previous line is the gravitational force on the object, that is, it's weight. So when x = R (we are at the surface of the Earth), F = mg (what you would normally use for weight at the surface of the Earth).

All the questions on gravity I’ve seen start like this, so when u learn this it will be an easy 3 marks.👊🏾

"The magnitude of this force is inversely proportional to $x^2$"

...so $P \propto \dfrac{1}{x^2}$ hence $P= \dfrac{k}{x^2}$ hence $ma=\dfrac{k}{x^2}$.

Now use the fact that "At the surface of the Earth the acceleration due to gravity is $g$. The Earth is modelled as a sphere of radius $R$" to determine $k$.

why wouldn't I treat this as a projectile and use Suvat? and why have they used double dot instead of a normal a in the mark scheme?

how would I do b? thanks

why wouldn't I treat this as a projectile and use Suvat? and why have they used double dot instead of a normal a in the mark scheme?